Epsilon nought

Description

Ordinal exponentiation is defined by $\omega^0 = 1$, $\omega^{\alpha + 1} = \omega^\alpha \cdot \omega$, and $\omega^\lambda = \sup_{\alpha < \lambda} \omega^\alpha$ for a limit ordinal $\lambda$. An epsilon number is an ordinal $\varepsilon$ with $\omega^\varepsilon = \varepsilon$. The ordinal $\varepsilon_0$, read "epsilon nought", is the least epsilon number. It is the limit of the sequence $$\omega, \quad \omega^\omega, \quad \omega^{\omega^\omega}, \quad \dots,$$ that is, $\varepsilon_0 = \sup \set{\omega_n : n \in \N}$ with $\omega_0 = \omega$ and $\omega_{n+1} = \omega^{\omega_n}$.

Check: let $\varepsilon = \sup_n \omega_n$. The map $\alpha \mapsto \omega^\alpha$ is continuous at limits, so $\omega^\varepsilon = \sup_n \omega^{\omega_n} = \sup_n \omega_{n+1} = \varepsilon$. If $\beta$ is an epsilon number, then $\omega \le \beta$, and induction on $n$ gives $\omega_n \le \beta$, so $\varepsilon \le \beta$. Since each $\omega_n$ is a countable ordinal, $\varepsilon_0$ is a countable union of countable sets, hence countable.